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Problem of the Week
Problem C and Solution
A Crafty Little Problem
Problem
A knot is tied near one end of a piece of string and beads are strung to form a necklace. The
beads are placed on the string in the following sequence: 1 red, 1 green, 1 blue, 2 red, 2 green,
2 blue, 3 red, 3 green, 3 blue, with the number of each colour increasing by one every time a
new group of beads is placed. The diagram illustrates how the first 18 beads are strung. How
many of the first 160 beads are blue?
R
G
B
R
R
B
G G
G G
B B R R R G
B
B
Solution
An equal number of red, green and blue beads occur after
3(1) = 3 beads are placed,
3(1) + 3(2) = 3 + 6 = 9 beads are placed,
3(1) + 3(2) + 3(3) = 3 + 6 + 9 = 18 beads are placed, and so on.
The greatest total that can be placed with equal numbers of red, green and blue
beads is 3(1) + 3(2) + 3(3) + · · · + 3(9) = 3 + 6 + 9 + · · · + 27 = 135. At this
point there are 135 ÷ 3 = 45 beads of each colour. We are at a point where we
have strung 9 red beads, 9 green beads and 9 blue beads in the last three groups.
The next groups will have 10 in them if there are enough beads. There are
160 − 135 = 25 beads left to place. We are able to place 10 red beads leaving 15
beads left to place. At this point there are 55 red beads. We can place 10 green
beads leaving 5 beads left to place. At this point there are 55 green beads. The
five remaining beads must be blue making the total number of blue beads 50.
∴ there are 50 blue beads on the necklace.